2. A small lake of volume 100 cubic meters has accidentally been polluted by 10 000 kilograms of a highly toxic substance. A river flows into and out of the lake at a rate of 20 000 cubic meters per hour. Assuming that the entering river contains fresh water and that the toxic substance is fully mixed throughout the lake at all times. find the number of hours for the mass of the pollutant to decrease to an acceptable 100 kilograms.

To solve this problem, we have to use the concept of differential equations and modeling the rate of change of pollutant concentration in the lake over time. Let's denote:

- $C(t)$ as the concentration of the pollutant in the lake at time $t$ (in kilograms per cubic meter).
- $V$ as the volume of the lake (given as ${10}^6$ cubic meters).
- $R_{in}$ as the rate at which the river flows into the lake (given as $20,000$ cubic meters per hour).
- $M(t)$ as the mass of the pollutant in the lake at time $t$ (in kilograms).

We want to find the time it takes for the mass of the pollutant ($M(t)$) to decrease to 100 kilograms, so we're looking for the time when $M(t)=100$.

The change in mass of the pollutant in the lake over time is given by the rate of inflow of pollutant minus the rate of outflow of pollutant:

$$\frac{dM}{dt}=R_{in}\cdot C_{in}\cdot V-R_{out}\cdot C(t)\cdot V$$

Where:
- $C_{in}$ is the concentration of the pollutant in the river, which is initially 10,000 kg / 20,000 m^3 = 0.5 kg/m^3.

We also know that $M(t)=C(t)\cdot V$, so we can rewrite the equation as:

$$\frac{dM}{dt}=R_{in}\cdot C_{in}\cdot V-R_{out}\cdot \frac{M(t)}{V}$$

Now, we can substitute the given values into the equation:

$$\frac{dM}{dt}=20,000\cdot 0.5\cdot {10}^6-20,000\cdot \frac{M(t)}{{10}^6}$$

Simplify the equation:

$$\frac{dM}{dt}={10}^6-20M(t)$$

Now, we have a first-order differential equation. We can solve it by separating variables:

$$\frac{dM}{{10}^6-20M(t)}=dt$$

Integrate both sides:

$$-\frac{1}{20}\ln |{10}^6-20M(t)|=t+C_1$$

Where $C_1$ is the constant of integration.

Now, we can solve for $M(t)$:

$$|{10}^6-20M(t)|=e^{-20(t+C_1)}$$

Since $M(t)$ cannot be negative (negative mass doesn't make sense), we can drop the absolute value:

$${10}^6-20M(t)=e^{-20(t+C_1)}$$

Now, we can solve for $M(t)$ when $M(t)=100$ (the acceptable level):

$${10}^6-20(100)=e^{-20(t+C_1)}$$

Simplify:

$${10}^6-2000=e^{-20(t+C_1)}$$

$$e^{20(t+C_1)}={10}^6-2000$$

$$t+C_1=\frac{\ln ({10}^6-2000)}{20}$$

Now, we can find the time $t$ it takes for $M(t)$ to decrease to 100 kilograms by substituting $M(t)=100$ and solving for $t$:

$$t=\frac{\ln ({10}^6-2000)}{20}-C_1$$